Group 2,7 and Redox

Question 1

State and explain the following trend for group 2 elements; atomic radius, first ionisation energy, melting point

Answer 1

Atomic radius increases down the group as the number of shells/energy levels increase. First ionisation energy decreases down the group as shielding increases so the attraction between the nucleus and outer electron decreases. Melting point decreases (except magnesium) because the size of the positive ions increases, and this creates weaker attraction to the delocalised electrons (i.e. weaker metallic bonding).

Question 2

Explain why group 2 metals have a smaller atomic radius than the preceding group 1 metal

Answer 2

The group 2 metals have a greater nuclear charge/more protons than the previous group 1 metal, but shielding remains the same, so there is stronger attraction between the nucleus and outer electrons. This draws them in slightly closer.

Question 3

Explain why the second ionisation energy of a group 2 metal is higher than its first ionisation energy

Answer 3

The electron is being removed from a positive ion so there is greater attraction to the electron being removed.

Question 4

Explain why the third ionisation energy of a group 2 metal is much higher than its second ionisation energy

Answer 4

The third electron is removed from a shell that is closer to the nucleus (as well as being an increasing positive ion). This means that there is even more attraction between the nucleus and electron being removed.

Question 5

Describe how titanium is extracted from its ore. Write two equations in your answer.

Answer 5

The main titanium ore contains titanium(IV) oxide, TiO2.

The titanium oxide is first reacted with chlorine and carbon to form titanium chloride, TiCl4, and carbon monoxide.

TiO2 + 2Cl2 + 2C à TiCl4 + 2CO

The titanium chloride is then reduced to titanium by reaction with magnesium.

TiCl4 + 2Mg à 2MgCl2 + Ti

Question 6

Write an equation to show how the following group 2 metals react with water (and steam for Mg) and state the pH of resulting solution; magnesium, calcium, strontium,barium

Answer 6

Mg(s) + 2H2O(l) à Mg(OH)2(s) + H2(g)

It produces a solution of pH 10

Magnesium with steam:

Mg(s) + H2O(g) à MgO(s)+ H2(g)

Ca(s) + 2H2O(l) à Ca(OH)2(aq) + H2(g)

It produces a solution of pH 11.

Sr(s) + 2H2O(l) à Sr(OH)2(aq) + H2(g)

It produces a colourless solution of pH 12.

Ba(s) + 2H2O(l) à Ba(OH)2(aq) + H2(g)

It produces a colourless solution of pH 13.

Question 7

What and explain the trend in reactivity of group 2 metals

Answer 7

Reactivity increases down the group as the atomic radius/shielding increases so there is less attraction between the nucleus and outer electrons. This means it becomes easier to lose an electron and react.

Question 8

What is the trend in the solubility of group 2 hydroxides?

Answer 8

The group 2 hydroxides become more soluble as you go down the group.

Question 9

What are two uses of group 2 hydroxides

Answer 9

Magnesium hydroxide (milk of magnesia) is used to treat indigestion or as a laxative. Calcium hydroxide (slaked lime) is used to neutralise acidic soil.

Question 10

What is the trend in the solubility of group 2 sulfates

Answer 10

The group 2 sulfates become less soluble as you go down the group.

Question 11

Describe the chemical test for sulfate ions

Answer 11

Add barium chloride/nitrate. A write precipitate of barium sulfate is formed if sulfate ions are present. Ba2+(aq) + SO42-(aq) à BaSO4(s)

Question 12

State and explain the medical use of barium sulfate

Answer 12

Barium sulfate (barium meal) can be shallowed to show the digestive tract under x-ray

Question 13

What are the states and appearances of the group 7 elements at room temperature?

Answer 13

F2 = pale green gas, Cl2 = dense green gas, Br2 = volatile orange liquid, I2 = grey solid, purple vapours when heated.

Question 14

Describe the structure and bonding of the group 7 elements.

Answer 14

Covalent bond, simple covalent molecules.

Question 15

State and explain the following trends of the group 7 elements; atomic radius, melting/boiling points, electronegativity, first ionisation energy.

Answer 15

Atomic radius increases down the group as the number of shells/energy levels increases. Electronegativity decreases down the group as the bonding pair of electrons becomes further from the nucleus and experience less attraction. First ionisation energy decreases down the group as the outer electrons from the nucleus becomes further from away/experiences less shielding so becomes easier to lose.

Question 16

Write a balanced equation for a displacement reaction involving a halogen and a halide. Turn this equation into an ionic equation

Answer 16

Cl2(aq) + 2KBr(aq) à 2KCl(aq) + Br2(aq) (yellow solution)

Cl2(aq) + 2Br-(aq) à 2Cl-(aq) + Br2(aq)

Cl2(aq) + 2KI(aq) à 2KCl(aq) + I2(aq) (brown solution)

Cl2(aq) + 2I-(aq) à 2Cl-(aq) + I2(aq)

Br2(aq) + 2KI(aq) à 2KBr(aq) + I2(aq) (brown solution)

Br2(aq) + 2I-(aq) à 2Br-(aq) + I2(aq)

Question 17

State and explain the trend in oxidising ability of the group 7 elements

Answer 17

Oxidising ability of the halogens decreases down the group. This is because the halogens must gain electrons when they react. This becomes harder as you go down the group because the number of shells/shielding/atomic radius increases so the electron is further away from the nucleus and experience less attration.

Question 18

List the observations for the reactions of chlorine/bromine/iodine water with potassium chloride/bromide/iodide

Answer 18

There is no reaction with iodine water.

Question 19

Explain why chlorine is added to drinking water

Answer 19

To sterilise it/kill pathogens/bacteria.

Question 20

Write an equation for the addition of chlorine to water. Which product kills bacteria

Answer 20

CL2 + H2O > HCLO + HCL

8. chlorate ions (ClO-) kills pathogens as it is a powerful oxidising agent.

Question 21

Write two equations for the addition of chlorine to water in the presence of sunlight.

Question 22

What is the main chemical in household bleach? Write an equation for its formation from chlorine

Answer 22

2NaOH(aq) + Cl2(g) à NaCl(aq) + NaClO(aq) + H2O(l). The chlorate ions (ClO-) is the active ingredient as it is a powerful oxidising agent.

Question 23

Write a half equation to show how chlorate(I) behaves as an oxidising agent

Answer 23

ClO- + H2O + 2e- à Cl- + 2OH-

Question 24

State and explain the trend in reducing ability of the halides

Answer 24

The halides become stronger reducing agents as you go down the group. This is because they lose electrons when they react. This becomes easier as they have more shells/energy levels/shielding as there is less attraction from the nucleus.

Question 25

Write an equation for the reaction of sodium chloride with sulfuric acid. State the observations made during this reaction

Answer 25

NaCl(s) + H2SO4(l) à NaHSO4(s) + HCl(g). Misty fumes of HCl

Question 26

Write two equations for the reaction of sodium bromide with sulfuric acid. State the observations made during these reactions

Answer 26

NaBr(s) + H2SO4(l) à NaHSO4(s) + HBr(g). Misty fumes of HBr

2H+ + 2Br- + H2SO4 → SO2 + 2H2O + Br2 . Brown fumes of Br2

Question 27

Write two equations for the reaction of sodium iodide with sulfuric acid. State the observations made during these reactions

Answer 27

NaI(s) + H2SO4(l) à NaHSO4(s) + HI(g). Misty fumes of HI.

8H+ + 8I– + H2SO4 à H2S + 4H2O + 4I2. Purple fumes/black ppt of I2, Colourles gas with bad smell of H2S. Colourless gas with choking odour of SO2, Yellow precpitate of S

Question 28

Describe the chemical test for halide ions (including use of ammonia)

Answer 28

First add silver nitrate and nitric acid. The silver ions form precipitates of silver halides (i.e. Ag+(aq) + I-(aq) à AgI(s)). AgCl is a white precipitate, AgBr is a cream precipitate and AgI is a yellow precipitate. Dilute ammonia will dissolve AgCl (AgCl(s) + 2NH3(aq) → Ag(NH3)2+(aq) + Cl−(aq) ) and concentrated ammonia will dissolve AgBr (AgBr(s) + 2NH3(aq) → Ag(NH3)2+(aq) + Br−(aq) )

Question 29

State and explain which type of acid is used in the chemical test in Q16

Answer 29

Nitric acid must be used as both hydrochloric acid and sulfuric acid will create precipitates with the silver ions.

Question 30

Halide ions using sulfuric acid

Question 31

Halide ions using silver nitrate

Question 32

Carbonate ions

Answer 32

Carbonate ions: add a strong acid. Effervesce of CO2 will be observed (2H+(aq) + CO32-(aq) à CO2(g) + H2O(l)). If the CO2 is bubbled through limewater, it turns cloudy/milky.

Question 33

Sulfate ions

Answer 33

Sulfate ions: Add barium chloride or barium nitrate. A white precipitate of barium sulfate is formed. Ba2+ (aq) + SO42- (aq) à BaSO4(s)

Question 34

Hydroxide ions

Answer 34

Hydroxide ions: use red litmus paper (turns blue) or Universal Indicator paper (turns blue/purple)

Question 35

Ammonium ions

Answer 35

Ammonium ions: add sodium hydroxide to turn ammonium ions into ammonia gas: NH4+(aq) + OH-(aq) à NH3(g) + H2O(l). Test the ammonia gas by using damp red litmus paper (turns blue) or damp Universal Indicator paper (turns blue/purple).

Question 36

Group 2 ions (using sodium hydroxide)

Answer 36

Group 2 ions (using sodium hydroxide): Group 2 hydroxides are formed i.e. Mg2+ (aq) + 2OH-(aq) à Mg(OH)2(s). Beryllium and magnesium hydroxide are insoluble so form white precipitates, the rest are soluble so form colourless solutions.

Question 37

Group 2 ions (using sulfuric acid)

Answer 37

Group 2 ions (using sulfuric acid): Group 2 sulfates are formed i.e. Ba2+ (aq) + SO42- (aq) à BaSO4(s). Calcium, strontium and barium sulfates are insoluble so form white precipitates, the rest are soluble so form colourless solutions